hdu2609(hash+STL+字典树)

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1040    Accepted Submission(s): 406

Problem Description

Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell me
How many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some).
For example 0110 express a necklace, you can rotate it. 0110 -> 1100 -> 1001 -> 0011->0110.

 

Input

The input contains multiple test cases.
Each test case include: first one integers n. (2<=n<=10000)
Next n lines follow. Each line has a equal length character string. (string only include '0','1').

 

Output

For each test case output a integer , how many different necklaces.

 

Sample Input

4011011001001001141010010110000001

 

Sample Output

12

 

Author

yifenfei

 

Source

奋斗的年代

 

Recommend

yifenfei

 

本题要求不同构的字符串的个数。

两个字符串同构是指其中一个可以仅经过循环移位得到另外一个。而同构可以用最小表示法来解决。

每读入一个字符串，就找到它的最小表示形式；在判断该最小表示形式是否已经出现过：若没有则计数器加1,并将该最小表示形式存储起来；若有则直接处理下一个。最终计数器即为所求。判断是否出现过可以hash表，但字符串的hash表不好维护，可以考虑用字典树判重或用STL中的set容器。

#include<iostream>#include<set>#include<string>using namespace std;//返回母串的最小子串的起始位置int minpresent(string &str){int i,j,k,len=str.size();i=0,j=1,k=0;while(i<len&&j<len&&k<len){if(str[(i+k)%len]==str[(j+k)%len])k++;else {if(str[(i+k)%len]>str[(j+k)%len])i=i+k+1;    else j=j+k+1;if(i==j)j++;k=0;}}return ++i<++j?i:j;}int main(){int n,i,Minid,ans;bool flag;string tmp,str;set<string>Set;while(~scanf("%d",&n)){Set.clear();ans=0;for(i=0;i<n;i++){cin>>tmp;Minid=minpresent(tmp)-1;str.assign(tmp,Minid,tmp.size()-Minid);if(Minid)str.append(tmp,0,Minid);if(!Set.count(str)){ans++;Set.insert(str);}}printf("%d\n",ans);}return 0;}